Oh, absolutely. I can do that. Let's take \(f_\pi(z) = e^{-z}-1\).
Take the function:
\[
\vartheta_\pi(x,z) = \sum_{n=0}^\infty f_\pi^{\circ n}(z) \frac{(-x)^n}{n!}\\
\]
When I write:
\[
\vartheta_\pi(ix,z) + \vartheta_\pi(-ix,z) = 2 Q(x,z)\\
\]
This is what I mean...
\[
\vartheta_\pi(ix,z) = \sum_{n=0}^\infty f_\pi^{\circ 2n}(z) \frac{(-1)^n x^{2n}}{2n!} - i \sum_{n=0}^\infty f_\pi^{\circ 2n+1}(z) \frac{(-1)^n x^{2n+1}}{2n+1!}\\
\]
Which is just done by collecting even and odd powers...
We only want the first sum, and we want to ignore the second part. This is because the first part, looks exactly like \(\cos(x)\), but the coefficients are \(f_\pi^{\circ 2n}(z)\).
Well...
\[
\vartheta_\pi(-ix,z) = \sum_{n=0}^\infty f_\pi^{\circ 2n}(z) \frac{(-1)^n x^{2n}}{2n!} + i \sum_{n=0}^\infty f_\pi^{\circ 2n+1}(z) \frac{(-1)^n x^{2n+1}}{2n+1!}\\
\]
So that:
\[
\vartheta_\pi(ix,z) + \vartheta_\pi(-ix,z) = 2 \sum_{n=0}^\infty f_\pi^{\circ 2n}(z) \frac{(-1)^n x^{2n}}{2n!}
\]
But the equation on the right is differintegrable; because \((f_\pi \circ f_\pi)^{\circ s}\) is Mellin transformable, and Attached with the Gamma expression I wrote above, it still is. So this means the left hand side is differintegrable.
So this proves that \(f_\pi^{\circ s}(z)\) does exist, but it isn't differintegrated in a conventional manner. We obviously need to rigorify this shit some more, but the general motions look like this.
To relate this to Euler's formula... by collecting even and odd powers again...
\[
e^{ix} = \sum_{n=0}^\infty \frac{(-1)^nx^{2n}}{2n!} + i \sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{2n+1!}\\
\]
Which famously reduces into:
\[
e^{ix} = \cos(x) + i \sin(x)\\
\]
Imagine we are doing the same thing, but we are doing something slightly different.
Let:
\[
Hu = u((f_\pi \circ f_\pi)^{\circ 1/2}) = u(h(z))\\
\]
Which is a linear operator. It takes functions holomorphic \(u(z)\) to the function \(Hu = u(h(z))= u((f_\pi \circ f_\pi)^{\circ 1/2}(z))\). (Again, you have to note that \(h = (f_\pi \circ f_\pi)^{\circ 1/2} \neq f_\pi\), the left element has \(1\) as a multiplier at \(0\), and the right has \(-1\) as a multiplier at \(0\))... This is the exact same equation as \(-1 = \sqrt{(-1)(-1)} = 1\). We have to pay attention how we take the root; and be very careful.
To justify it's a linear operator. Let \(\alpha, \beta \in \mathbb{C}\). Let \(u,v : \mathbb{C} \to \mathbb{C}\). Then:
\[
H\left(\alpha u(z) + \beta v(z)\right) = \alpha u(h(z)) + \beta v(h(z)) = \alpha Hu + \beta Hv\\
\]
Then:
\[
\cos(Hx)z = \sum_{n=0}^\infty H^{2n}z \frac{(-1)^n x^{2n}}{2n!} = \sum_{n=0}^\infty f_\pi^{\circ 2n}(z) \frac{(-1)^n x^{2n}}{2n!}\\
\]
This function is differintegrable in \(x\), because:
\[
e^{Hx}z = \sum_{n=0}^\infty H^n \frac{x^n}{n!}\\
\]
is differintegrable; and by a change of variables I described with the \(\Gamma\) function, we know that both:
\[
\cos(Hx)z + i\sin(Hx)z = e^{iHx}z\\
\]
are differintegrable. So now, we're only going to focus on \(\cos\) (we could focus on \(\sin\) just as well, but \(\cos\) is easier). If I make a different operator \(P\), such that:
\[
Pu = u(f_\pi(z))\\
\]
The function:
\[
\cos(Px)z = \cos(Hx)z\\
\]
So that, even though \(e^{-Hx} \neq e^{-Px} = \vartheta_\pi(x)\)--and \(e^{Px}\) is only differintegrable in a distributional sense (where it is the solution of an algebraic equation extending the integral transform), we can successfully take:
\[
\frac{d^{s}}{dx^{s}} e^{Px}\\
\]
Where, all we have to do is take:
\[
\frac{d^s}{dx^s} \frac{e^{iPx} + e^{-iPx}}{2} = \frac{d^s}{dx^s} \cos(Px) = \frac{d^s}{dx^s} \cos(Hx)\\
\]
Out pops a bunch of Gamma nonsense that looks like something from Riemann's journal (Again, I'm a big analytic number theory guy, lol). Exactly what pops out, is what you saw above (I may have fudged some subtleties, this stuff still gives me a headache, but it's something like this). But in the distributional sense:
\[
\frac{d^s}{dx^s} \frac{e^{iPx} + e^{-iPx}}{2} = P^s \frac{i^s + i^{-s}}{2} = P^s \frac{e^{i\pi s/2} + e^{-i \pi s/2}}{2} = P^s \cos(\pi s /2) = H^s \frac{\Gamma(s/2)\Gamma(1-s/2)}{\Gamma(1-s)}\\
\]
I expect \(P^s = \frac{d^s}{dx^s} e^{Px} = \frac{d^s}{dx^s} \vartheta_\pi(x)\) to actually be pretty poorly behaved. If it was super well behaved, then \(\vartheta_\pi\) wouldn't need distributional analysis. So \(P^s\) may have singularities; or malaligned behaviour. I'm going to have to look at it much deeper. I'm just giving a sketch for now.
So essentially, the point is, under the inverse Mellin transform as a distributional transform (distributional Fourier transform), we have the equality:
\[
P^s \cos(\pi s /2) = H^s \frac{\Gamma(s/2)\Gamma(1-s/2)}{\Gamma(1-s)}\\
\]
So even though the Mellin transform of \(\vartheta_\pi\) doesn't converge; some extension of the Mellin transform of \(\vartheta_\pi\) does (advanced functional analysis shit, I'm going to have to put on my thinking cap). And the converged version is the above formula (forgoing I didn't forget a negative sign here or there in my analysis
)
Regards, James
EDIT:
Please remember that \(P\) and \(H\) are linear operators. But additionally they are actionable under a Fourier like transform. Additionally \(P^2 = H^2\). This means, we can construct a Hilbert space (In our case it's just looks like a Hardy space). We can look at these things like Gottfried (Heisenberg), and \(P,H\) are just infinite square matrices acting on an infinite vector space. We can look at this like waves, and that we are altering frequencies of the waves through \(P,H\)... Schrodinger shit. Or we can collectively say all of us are Von Neumann, and we're going to hands down prove both of these interpretations are the same
.
Let's do some Von Neumann. I believe I can generate an asymptotic series at \(z\approx 0\) for \(f_\pi^{\circ 1/2}(z) =g(z)\) which starts with the leading term \(g(z) = iz + O(z^2)\). Just as \(\sqrt{-1} = i\)
The choice is ours.
Take the function:
\[
\vartheta_\pi(x,z) = \sum_{n=0}^\infty f_\pi^{\circ n}(z) \frac{(-x)^n}{n!}\\
\]
When I write:
\[
\vartheta_\pi(ix,z) + \vartheta_\pi(-ix,z) = 2 Q(x,z)\\
\]
This is what I mean...
\[
\vartheta_\pi(ix,z) = \sum_{n=0}^\infty f_\pi^{\circ 2n}(z) \frac{(-1)^n x^{2n}}{2n!} - i \sum_{n=0}^\infty f_\pi^{\circ 2n+1}(z) \frac{(-1)^n x^{2n+1}}{2n+1!}\\
\]
Which is just done by collecting even and odd powers...
We only want the first sum, and we want to ignore the second part. This is because the first part, looks exactly like \(\cos(x)\), but the coefficients are \(f_\pi^{\circ 2n}(z)\).
Well...
\[
\vartheta_\pi(-ix,z) = \sum_{n=0}^\infty f_\pi^{\circ 2n}(z) \frac{(-1)^n x^{2n}}{2n!} + i \sum_{n=0}^\infty f_\pi^{\circ 2n+1}(z) \frac{(-1)^n x^{2n+1}}{2n+1!}\\
\]
So that:
\[
\vartheta_\pi(ix,z) + \vartheta_\pi(-ix,z) = 2 \sum_{n=0}^\infty f_\pi^{\circ 2n}(z) \frac{(-1)^n x^{2n}}{2n!}
\]
But the equation on the right is differintegrable; because \((f_\pi \circ f_\pi)^{\circ s}\) is Mellin transformable, and Attached with the Gamma expression I wrote above, it still is. So this means the left hand side is differintegrable.
So this proves that \(f_\pi^{\circ s}(z)\) does exist, but it isn't differintegrated in a conventional manner. We obviously need to rigorify this shit some more, but the general motions look like this.
To relate this to Euler's formula... by collecting even and odd powers again...
\[
e^{ix} = \sum_{n=0}^\infty \frac{(-1)^nx^{2n}}{2n!} + i \sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{2n+1!}\\
\]
Which famously reduces into:
\[
e^{ix} = \cos(x) + i \sin(x)\\
\]
Imagine we are doing the same thing, but we are doing something slightly different.
Let:
\[
Hu = u((f_\pi \circ f_\pi)^{\circ 1/2}) = u(h(z))\\
\]
Which is a linear operator. It takes functions holomorphic \(u(z)\) to the function \(Hu = u(h(z))= u((f_\pi \circ f_\pi)^{\circ 1/2}(z))\). (Again, you have to note that \(h = (f_\pi \circ f_\pi)^{\circ 1/2} \neq f_\pi\), the left element has \(1\) as a multiplier at \(0\), and the right has \(-1\) as a multiplier at \(0\))... This is the exact same equation as \(-1 = \sqrt{(-1)(-1)} = 1\). We have to pay attention how we take the root; and be very careful.
To justify it's a linear operator. Let \(\alpha, \beta \in \mathbb{C}\). Let \(u,v : \mathbb{C} \to \mathbb{C}\). Then:
\[
H\left(\alpha u(z) + \beta v(z)\right) = \alpha u(h(z)) + \beta v(h(z)) = \alpha Hu + \beta Hv\\
\]
Then:
\[
\cos(Hx)z = \sum_{n=0}^\infty H^{2n}z \frac{(-1)^n x^{2n}}{2n!} = \sum_{n=0}^\infty f_\pi^{\circ 2n}(z) \frac{(-1)^n x^{2n}}{2n!}\\
\]
This function is differintegrable in \(x\), because:
\[
e^{Hx}z = \sum_{n=0}^\infty H^n \frac{x^n}{n!}\\
\]
is differintegrable; and by a change of variables I described with the \(\Gamma\) function, we know that both:
\[
\cos(Hx)z + i\sin(Hx)z = e^{iHx}z\\
\]
are differintegrable. So now, we're only going to focus on \(\cos\) (we could focus on \(\sin\) just as well, but \(\cos\) is easier). If I make a different operator \(P\), such that:
\[
Pu = u(f_\pi(z))\\
\]
The function:
\[
\cos(Px)z = \cos(Hx)z\\
\]
So that, even though \(e^{-Hx} \neq e^{-Px} = \vartheta_\pi(x)\)--and \(e^{Px}\) is only differintegrable in a distributional sense (where it is the solution of an algebraic equation extending the integral transform), we can successfully take:
\[
\frac{d^{s}}{dx^{s}} e^{Px}\\
\]
Where, all we have to do is take:
\[
\frac{d^s}{dx^s} \frac{e^{iPx} + e^{-iPx}}{2} = \frac{d^s}{dx^s} \cos(Px) = \frac{d^s}{dx^s} \cos(Hx)\\
\]
Out pops a bunch of Gamma nonsense that looks like something from Riemann's journal (Again, I'm a big analytic number theory guy, lol). Exactly what pops out, is what you saw above (I may have fudged some subtleties, this stuff still gives me a headache, but it's something like this). But in the distributional sense:
\[
\frac{d^s}{dx^s} \frac{e^{iPx} + e^{-iPx}}{2} = P^s \frac{i^s + i^{-s}}{2} = P^s \frac{e^{i\pi s/2} + e^{-i \pi s/2}}{2} = P^s \cos(\pi s /2) = H^s \frac{\Gamma(s/2)\Gamma(1-s/2)}{\Gamma(1-s)}\\
\]
I expect \(P^s = \frac{d^s}{dx^s} e^{Px} = \frac{d^s}{dx^s} \vartheta_\pi(x)\) to actually be pretty poorly behaved. If it was super well behaved, then \(\vartheta_\pi\) wouldn't need distributional analysis. So \(P^s\) may have singularities; or malaligned behaviour. I'm going to have to look at it much deeper. I'm just giving a sketch for now.
So essentially, the point is, under the inverse Mellin transform as a distributional transform (distributional Fourier transform), we have the equality:
\[
P^s \cos(\pi s /2) = H^s \frac{\Gamma(s/2)\Gamma(1-s/2)}{\Gamma(1-s)}\\
\]
So even though the Mellin transform of \(\vartheta_\pi\) doesn't converge; some extension of the Mellin transform of \(\vartheta_\pi\) does (advanced functional analysis shit, I'm going to have to put on my thinking cap). And the converged version is the above formula (forgoing I didn't forget a negative sign here or there in my analysis
)Regards, James
EDIT:
Please remember that \(P\) and \(H\) are linear operators. But additionally they are actionable under a Fourier like transform. Additionally \(P^2 = H^2\). This means, we can construct a Hilbert space (In our case it's just looks like a Hardy space). We can look at these things like Gottfried (Heisenberg), and \(P,H\) are just infinite square matrices acting on an infinite vector space. We can look at this like waves, and that we are altering frequencies of the waves through \(P,H\)... Schrodinger shit. Or we can collectively say all of us are Von Neumann, and we're going to hands down prove both of these interpretations are the same
.Let's do some Von Neumann. I believe I can generate an asymptotic series at \(z\approx 0\) for \(f_\pi^{\circ 1/2}(z) =g(z)\) which starts with the leading term \(g(z) = iz + O(z^2)\). Just as \(\sqrt{-1} = i\)
The choice is ours.